Past Exam Questions

(i) We use the standard normal distribution to find the mean \(\mu\) and standard deviation \(\sigma\).

For lengths less than 6 cm: \(Z = \frac{6 - \mu}{\sigma} = -1.253\)

For lengths more than 12 cm: \(Z = \frac{12 - \mu}{\sigma} = 0.648\)

Solving these equations simultaneously gives:

\(\mu = 9.9\)

\(\sigma = 3.15 \text{ or } 3.16\)

(ii) We need \(P(Z < -1 \text{ or } Z > 1)\).

This is equivalent to \(1 - \Phi(1) + \Phi(-1)\).

\(= 2 - 2 \times 0.8413\)

\(= 0.3174\)

In a sample of 1000, the expected number is \(0.3174 \times 1000 = 317\).

(i) Given that the mean μ is three times the standard deviation σ, we have μ = 3σ. The probability P(X < 25) = 0.648 corresponds to a standard normal variable z = 0.38. Using the standardization formula:

\(z = \frac{25 - \mu}{\sigma} = 0.38\)

Substitute \(\mu = 3\sigma\) into the equation:

\(\frac{25 - 3\sigma}{\sigma} = 0.38\)

Solving for \(\sigma\):

\(25 - 3\sigma = 0.38\sigma\)

\(25 = 3.38\sigma\)

\(\sigma = \frac{25}{3.38} \approx 7.40\)

Then, \(\mu = 3\sigma = 3 \times 7.40 = 22.2\).

(ii) The probability that a single value of X is greater than 25 is 1 - 0.648 = 0.352. We need the probability that exactly 4 out of 6 values are greater than 25. This follows a binomial distribution:

\(P(4) = \binom{6}{4} \times (0.352)^4 \times (0.648)^2\)

\(P(4) = 15 \times (0.0152) \times (0.419904)\)

\(P(4) \approx 0.0967\)

(ii) To find the proportion of winter days with a minimum temperature below zero, we standardize the variable:

\(z = \frac{0 - \mu}{2\mu} = -0.5\)

The probability \(P(z < -0.5) = 1 - 0.6915 = 0.309\) or 30.9%.

(iii) To find how many of the 70 days have a temperature more than three times the mean:

\(z = \frac{3\mu - \mu}{2\mu} = 1\)

The probability \(P(z > 1) = 1 - 0.8413 = 0.1587\).

Thus, the expected number of days is \(70 \times 0.1587 = 11.1\), so approximately 11 or 12 days.

(iv) Given the probability of the temperature being above 6 °C is 0.0735, we find \(\mu\):

\(z = 1.45\)

\(1.45 = \frac{6 - \mu}{2\mu}\)

Solving for \(\mu\), we get \(\mu = 1.54\).

(i) Given that 94% of the letters are within 12 g of the mean, we use the z-score for 47% (since 94% is the total for both tails, 47% for one tail) which is approximately 1.882. The equation is:

\(1.882 = \frac{32 - 20}{\sigma}\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{12}{1.882} \approx 6.38\)

(ii) To find the probability that a letter weighs more than 13 g, we standardize:

\(P(x > 13) = P\left( z > \frac{13 - 20}{6.38} \right)\)

\(= P(z > -1.0978)\)

Using the standard normal distribution table, \(P(z > -1.0978) = 0.864\).

(iii) The probability that a letter weighs more than 32 g is 0.03 (since 6% are more than 12 g above the mean). We use the binomial distribution for at least 2 out of 7:

\(P(\text{at least 2}) = 1 - P(0, 1)\)

\(= 1 - (0.97)^7 - (0.03)(0.97)^6 \times \binom{7}{1}\)

\(= 1 - (0.97)^7 - 7 \times (0.03)(0.97)^6\)

\(= 0.0171\)

(i) Given that 10% of the straws are shorter than 20 cm, we have \(P(x < 20) = 0.1\). Using the standard normal distribution, this corresponds to a \(z\)-value of approximately \(-1.282\).

Standardizing, \(z = \frac{20 - \mu}{0.8}\).

Thus, \(-1.282 = \frac{20 - \mu}{0.8}\).

Solving for \(\mu\), we get \(\mu = 20 + 1.282 \times 0.8 = 21.0256 \approx 21.0 \text{ cm}\).

(ii) We need to find \(P(21.5 < x < 22.5)\).

Standardizing, \(P\left( \frac{21.5 - 21.03}{0.8} < z < \frac{22.5 - 21.03}{0.8} \right)\).

This simplifies to \(P(0.5875 < z < 1.8375)\).

Using the standard normal distribution table, \(\Phi(1.8375) - \Phi(0.5875) = 0.9670 - 0.7217 = 0.2453\).

Now, we use the binomial distribution for 4 trials with probability \(p = 0.2453\).

We need \(P(X < 2) = P(0) + P(1)\).

\(P(0) = (0.7547)^4\).

\(P(1) = 4 \times (0.2453) \times (0.7547)^3\).

Calculating these gives \(P(0) + P(1) = 0.7547^4 + 4 \times 0.2453 \times 0.7547^3 = 0.746\).

(a) We start with the given equation 3μ = 7σ² and the probability P(X > 2μ) = 0.1016. Standardizing, we have:

\(z = \frac{2μ - μ}{σ} = \frac{μ}{σ}\)

\(From the standard normal distribution table, P(Z > 1.272) = 0.1016, so:\)

\(\frac{7σ}{3} = 1.272\)

Solving for σ, we get:

\(σ = 0.545\)

Substituting back to find μ:

\(μ = \frac{7σ^2}{3} = 0.693\)

(b) Given Y ~ N(33, 21), we need to find a such that P(33 − a < Y < 33 + a) = 0.5. This implies:

\(P(X < a + 33) = 0.75\)

\(From the standard normal distribution table, z = 0.674 for 0.75 probability. Standardizing:\)

\(\frac{a + 33 - 33}{\sqrt{21}} = 0.674\)

Solving for a gives:

\(a = 3.09\)

(i) Given that 79% of people have visits lasting less than 10 minutes, we find the z-score corresponding to 0.79, which is 0.807. Using the formula for the z-score:

\(z = \frac{X - \mu}{\sigma}\)

where \(X = 10\), \(\mu = 8.2\), and \(z = 0.807\). Solving for \(\sigma\):

\(0.807 = \frac{10 - 8.2}{\sigma}\)

\(\sigma = \frac{1.8}{0.807} \approx 2.23\)

(ii) We need to find \(P(|X - 8.2| > 1)\), which is equivalent to \(P(X > 9.2) + P(X < 7.2)\). Standardizing these values:

\(P\left(\left| z \right| > \frac{1}{2.23}\right)\)

\(P(|z| > 0.4484) = 2 \times (1 - \Phi(0.4484))\)

\(= 2 \times (1 - 0.6729) = 0.654\)

(iii) Let \(p = 0.21\) be the probability of a visit lasting longer than 10 minutes. We want \(P(X > 2)\) for a binomial distribution with \(n = 6\) and \(p = 0.21\):

\(P(X > 2) = 1 - P(X \leq 2)\)

\(= 1 - \{ (0.79)^6 + 6 \times (0.21) \times (0.79)^5 + 15 \times (0.21)^2 \times (0.79)^4 \}\)

\(= 1 - \{ 0.2621 + 0.2371 + 0.3888 \}\)

\(= 0.112\)

(i) To find the probability that each car runs out of fuel before travelling 367 km, we standardize the normal distribution for each car.

For the Zotoc car:

\(z = \frac{367 - 320}{21.6} = 2.176\)

The probability \(P(Z < 2.176)\) is approximately 0.985.

For the Ganmor car:

\(z = \frac{367 - 350}{7.5} = 2.267\)

The probability \(P(Z < 2.267)\) is approximately 0.988.

(ii) We are given that the probability that a Zotoc car can travel at least \(320 + d\) km is 0.409. This corresponds to a z-score of 0.23 (since \(P(Z > 0.23) = 0.409\)).

Standardizing, we have:

\(0.23 = \frac{x - 320}{21.6}\)

Solving for \(x\), we get:

\(x = 0.23 \times 21.6 + 320 = 324.968\)

Thus, \(d = 324.968 - 320 = 4.97\).

(i) We have two conditions:

1. 8 out of 10 children can jump more than 127 cm, so \(P(X > 127) = 0.8\). This corresponds to a z-score of \(-0.842\).

2. 1 out of 3 children can jump more than 135 cm, so \(P(X > 135) = 0.333\). This corresponds to a z-score of \(0.431\).

Using the z-score formula \(z = \frac{x - \mu}{\sigma}\), we set up the equations:

\(0.431 = \frac{135 - \mu}{\sigma}\)

\(-0.842 = \frac{127 - \mu}{\sigma}\)

Solving these equations simultaneously gives \(\mu = 132\) and \(\sigma = 6.29\).

(ii) To find the probability that a child will not be able to jump 145 cm, we calculate \(P(X < 145)\).

Standardize: \(z = \frac{145 - 132}{6.29} = 2.023\).

Using normal distribution tables, \(P(z < 2.023) = 0.978\).

(iii) Let \(p = \frac{1}{3}\) be the probability that a child can jump more than 135 cm.

We need \(P(\text{at least 2 out of 8}) = 1 - P(0, 1)\).

\(P(0, 1) = \left(\frac{2}{3}\right)^8 + \binom{8}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^7\).

Calculate: \(P(0, 1) = 0.195\).

Thus, \(P(\text{at least 2}) = 1 - 0.195 = 0.805\).

(i) To find P(X < 2μ), we standardize the variable:

\(P(X < 2μ) = P\left( z < \frac{2μ - μ}{σ} \right) = P\left( z < \frac{μ}{σ} \right).\)

Given 5σ = 3μ, we have \(\frac{μ}{σ} = \frac{5}{3}\).

\(Thus, P(X < 2μ) = P\left( z < \frac{5}{3} \right) = 0.952.\)

(ii) We are given P(X < \(\frac{1}{3}μ\)) = 0.8524. Standardizing gives:

\(P\left( z < \frac{\frac{1}{3}μ - μ}{σ} \right) = P\left( z < \frac{-2μ}{3σ} \right).\)

From the standard normal distribution table, \(\frac{-2μ}{3σ} = -1.047\).

Solving for μ, we get \(\frac{-2μ}{3σ} = -1.047\) which gives \(μ = -1.57σ\).

First, calculate the probability that a bag weighs more than 2.6 kg:

\(P(X > 2.6) = \frac{134}{5000} = 0.0268\)

The probability that a bag weighs less than 2.6 kg is:

\(P(X < 2.6) = 1 - 0.0268 = 0.9732\)

Using the standard normal distribution, find the z-score corresponding to this probability:

\(\frac{2.6 - 2.55}{\sigma} = 1.93\)

Solving for \(\sigma\):

\(\sigma = \frac{2.6 - 2.55}{1.93} = 0.0259\)

Given that \(X \sim N(\mu, \sigma^2)\), we have two probabilities:

1. \(P(X > 30.0) = 0.1480\)

2. \(P(X > 20.9) = 0.6228\)

Using the standard normal distribution, we convert these to \(z\)-scores:

For \(X = 30.0\):

\(z = \frac{30 - \mu}{\sigma}\)

From the standard normal table, \(P(Z > 1.045) = 0.1480\), so \(z = 1.045\).

Thus, \(\frac{30 - \mu}{\sigma} = 1.045\).

For \(X = 20.9\):

\(z = \frac{20.9 - \mu}{\sigma}\)

From the standard normal table, \(P(Z > -0.313) = 0.6228\), so \(z = -0.313\).

Thus, \(\frac{20.9 - \mu}{\sigma} = -0.313\).

We now have two equations:

1. \(30 - \mu = 1.045 \sigma\)

2. \(20.9 - \mu = -0.313 \sigma\)

Solving these simultaneously:

Subtract the second equation from the first:

\((30 - \mu) - (20.9 - \mu) = 1.045 \sigma + 0.313 \sigma\)

\(9.1 = 1.358 \sigma\)

\(\sigma = \frac{9.1}{1.358} = 6.70\)

Substitute \(\sigma = 6.70\) into the first equation:

\(30 - \mu = 1.045 \times 6.70\)

\(30 - \mu = 7.0015\)

\(\mu = 30 - 7.0015 = 23.0\)

Thus, \(\mu = 23.0\) and \(\sigma = 6.70\).

(i) To find the probability that a randomly chosen bar of soap weighs more than 128 grams, we standardize the variable:

\(P(X > 128) = P\left( z > \frac{128 - 125}{4.2} \right)\)

\(= P(z > 0.7143)\)

Using the standard normal distribution table, \(P(z > 0.7143) = 1 - 0.7623 = 0.238\).

(ii) We need to find the value of \(k\) such that \(P(k < X < 128) = 0.7465\).

\(P(X < 128) = 0.7465 + 0.2377 = 0.9842\)

Using the standard normal distribution table, \(z = -2.15\) corresponds to \(P(X < k)\).

\(-2.15 = \frac{k - 125}{4.2}\)

Solving for \(k\), we get \(k = 116\).

(iii) For five bars of soap, we find the probability that more than two weigh more than 128 grams:

\(P(X > 2) = P(3, 4, 5) = 1 - P(0, 1, 2)\)

Using the binomial distribution:

\(= \binom{5}{3} (0.2377)^3 (0.7623)^2 + \binom{5}{4} (0.2377)^4 (0.7623)^1 + \binom{5}{5} (0.2377)^5\)

\(= 0.07804 + 0.01216 + 0.0007588\)

\(= 0.0910\)

(i) The mean \(\mu\) is estimated as the median of the data, which is the middle value of the box-and-whisker plot. From the diagram, the median is 51 km h\(^{-1}\).

(ii) To estimate the standard deviation \(\sigma\), we use the interquartile range (IQR) and the properties of the normal distribution. The IQR is the difference between the upper quartile (63 km h\(^{-1}\)) and the median (51 km h\(^{-1}\)), which is 12 km h\(^{-1}\).

For a normal distribution, the IQR corresponds to approximately 1.34896 standard deviations (since \(z = \pm 0.674\) for the quartiles). Therefore, \(\sigma\) can be estimated by:

\(\sigma = \frac{63 - 51}{0.674} = \frac{12}{0.674} \approx 17.8\)

(i) To find the probability that a journey takes between 85 and 100 minutes, we standardize the values using the formula:

\(z = \frac{x - \mu}{\sigma}\)

For \(x = 85\), \(z = \frac{85 - 100}{7} = -2.143\).

The probability \(P(85 < x < 100) = 0.5 - P(z < -2.143)\).

Using the standard normal distribution table, \(P(z < -2.143) = 1 - \Phi(2.143) = 0.0161\).

Thus, \(P(85 < x < 100) = 0.5 - 0.0161 = 0.484\).

(ii) For 'standard' journeys, which are the middle 34%, we find the z-scores corresponding to the cumulative probabilities of 0.33 and 0.67.

Using \(z = \Phi^{-1}(0.67) = 0.44\).

For the upper limit, \(0.44 = \frac{a - 100}{7}\), solving gives \(a = 103.1\) minutes.

For the lower limit, \(-0.44 = \frac{a - 100}{7}\), solving gives \(a = 96.9\) minutes.

(i) We know that 225 out of 900 cartons contain more than 1002 millilitres. Therefore, the probability \(P(X > 1002) = \frac{225}{900} = 0.25\).

Using the standard normal distribution, \(P(Z > z) = 0.25\) corresponds to \(z = 0.674\) (using standard normal tables or calculator).

Standardizing, we have:

\(\frac{1002 - \mu}{8} = 0.674\)

Solving for \(\mu\):

\(1002 - \mu = 0.674 \times 8\)

\(1002 - \mu = 5.392\)

\(\mu = 1002 - 5.392 = 996.608\)

Rounding to the nearest whole number, \(\mu = 997\).

(ii) The probability that a carton contains more than 1002 millilitres is \(p = 0.25\). We want the probability that exactly 2 out of 3 chosen cartons contain more than 1002 millilitres.

This is a binomial probability problem with \(n = 3\), \(k = 2\), and \(p = 0.25\).

\(P(2) = \binom{3}{2} \times (0.25)^2 \times (0.75)^1\)

\(P(2) = 3 \times 0.0625 \times 0.75\)

\(P(2) = 0.140625\)

Rounding to three significant figures, the probability is 0.140.

The problem states that the daily minimum temperature follows a normal distribution \(N(\mu, 40.0)\). We are given that the probability of the temperature being above 0°C is 0.8888.

Using the standard normal distribution table, a probability of 0.8888 corresponds to a \(z\)-score of approximately 1.22. However, since we are dealing with the probability above 0°C, we use \(z = -1.22\).

The \(z\)-score formula is:

\(z = \frac{X - \mu}{\sigma}\)

where \(X = 0\) (the temperature), \(\mu\) is the mean, and \(\sigma = \sqrt{40}\).

Substitute the values into the formula:

\(-1.22 = \frac{0 - \mu}{\sqrt{40}}\)

Solving for \(\mu\):

\(-1.22 \times \sqrt{40} = -\mu\)

\(\mu = 1.22 \times \sqrt{40}\)

\(\mu \approx 7.72\)

(i) We know that 25% of infections clear up in less than 7 days, which corresponds to a z-score of approximately -0.674. Using the standardization formula:

\(z = \frac{X - \mu}{\sigma}\)

Substitute the known values:

\(-0.674 = \frac{7 - \mu}{2.6}\)

Solving for \(\mu\):

\(-0.674 \times 2.6 = 7 - \mu\)

\(-1.7524 = 7 - \mu\)

\(\mu = 7 + 1.7524 = 8.75\)

(ii) We need to find \(P(X > 6.2)\) where \(X\) is normally distributed with mean 6.5 and standard deviation 2.6. Standardizing:

\(P(X > 6.2) = P\left( Z > \frac{6.2 - 6.5}{2.6} \right)\)

\(= P(Z > -0.1154)\)

Using the symmetry of the normal distribution, \(P(Z > -0.1154) = 1 - P(Z < -0.1154)\). Since \(P(Z < -0.1154) \approx 0.454\),

\(P(Z > -0.1154) = 1 - 0.454 = 0.546\)

(i) To find the standard deviation \(\sigma\), we use the standard normal distribution formula:

\(z = \frac{X - \mu}{\sigma}\)

Given \(P(X > 5.5) = 0.0465\), we find the corresponding \(z\)-value from the standard normal distribution table, which is approximately 1.68. Thus,

\(z = \frac{5.5 - 4.5}{\sigma} = 1.68\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{1}{1.68} = 0.595\)

(ii) To find the probability that \(X\) lies between 3.8 and 4.8, we calculate the \(z\)-values for 3.8 and 4.8:

\(z_1 = \frac{3.8 - 4.5}{0.595} = -1.176\)

\(z_2 = \frac{4.8 - 4.5}{0.595} = 0.504\)

The probability is given by:

\(P(3.8 < X < 4.8) = \Phi(0.504) - (1 - \Phi(1.176))\)

Using the standard normal distribution table, \(\Phi(0.504) \approx 0.6929\) and \(\Phi(1.176) \approx 0.8802\), so:

\(P(3.8 < X < 4.8) = 0.6929 - (1 - 0.8802) = 0.573\)

(a) Let the mean be \(\mu = 2s\). The standard normal variable \(Z\) is given by:

\(Z = \frac{X - \mu}{s} = \frac{5.2 - 2s}{s}\)

Given \(P(X > 5.2) = 0.9\), we have \(P(Z > z) = 0.9\), which implies \(P(Z < z) = 0.1\). From standard normal tables, \(z = -1.282\).

Thus, \(\frac{5.2 - 2s}{s} = -1.282\).

Solving for \(s\):

\(5.2 - 2s = -1.282s\)

\(5.2 = 0.718s\)

\(s = \frac{5.2}{0.718} \approx 7.24 \text{ or } 7.23\)

(b) The probability that a value lies between \(\mu - \sigma\) and \(\mu + \sigma\) is \(P(|Z| < 1)\).

From standard normal tables, \(P(|Z| < 1) = 0.6826\).

Thus, the expected number of observations is:

\(0.6826 \times 800 = 546.08\)

Therefore, approximately 546 observations (accept 547).